int_0^1 frac{tan^{-1} x}{1 + x^2} ,dx = | vedclass

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Question

(D) Let $I = \int_0^1 \frac{	an^{-1} x}{1 + x^2} \,dx$. Substitute $t = 	an^{-1} x$. Then,$dt = \frac{1}{1 + x^2} \,dx$. When $x = 0$,$t = 	an^{-1}

Vedclass · Accepted Answer

$\frac{\pi^2}{32}$

Vedclass · Answer

(D) Let $I = \int_0^1 \frac{	an^{-1} x}{1 + x^2} \,dx$. Substitute $t = 	an^{-1} x$. Then,$dt = \frac{1}{1 + x^2} \,dx$. When $x = 0$,$t = 	an^{-1}(0) = 0$. When $x = 1$,$t = 	an^{-1}(1) = \frac{\pi}{4}$. Substituting these into the integral,we get: $I = \int_0^{\pi/4} t \,dt = \left[ \frac{t^2}{2} ight]_0^{\pi/4} = \frac{1}{2} \left( \left( \frac{\pi}{4} ight)^2 - 0^2 ight) = \frac{1}{2} \cdot \frac{\pi^2}{16} = \frac{\pi^2}{32}$.

$\int_0^1 \frac{\tan^{-1} x}{1 + x^2} \,dx = $

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